YES

We show the termination of the TRS R:

  g(s(x)) -> f(x)
  f(|0|()) -> s(|0|())
  f(s(x)) -> s(s(g(x)))
  g(|0|()) -> |0|()

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: g#(s(x)) -> f#(x)
p2: f#(s(x)) -> g#(x)

and R consists of:

r1: g(s(x)) -> f(x)
r2: f(|0|()) -> s(|0|())
r3: f(s(x)) -> s(s(g(x)))
r4: g(|0|()) -> |0|()

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: g#(s(x)) -> f#(x)
p2: f#(s(x)) -> g#(x)

and R consists of:

r1: g(s(x)) -> f(x)
r2: f(|0|()) -> s(|0|())
r3: f(s(x)) -> s(s(g(x)))
r4: g(|0|()) -> |0|()

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        g#_A(x1) = ((0,1),(0,0)) x1
        s_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        f#_A(x1) = ((1,1),(0,0)) x1
    
    2. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        g#_A(x1) = (3,3)
        s_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        f#_A(x1) = ((1,1),(1,1)) x1
    
    3. matrix interpretations:
    
      carrier: N^2
      order: standard order
      interpretations:
        g#_A(x1) = (2,3)
        s_A(x1) = ((1,1),(1,1)) x1 + (1,1)
        f#_A(x1) = ((0,1),(1,1)) x1
    

The next rules are strictly ordered:

  p1, p2

We remove them from the problem.  Then no dependency pair remains.